Question

Derive expressions for average energy of a body executing SHM.

Answer 1

Displacement x of a particle from mean position (equilibrium) is given by 
              x = A Sin ωt
A = amplitude of Simple Harmonic Motion.
ω = angular frequency
m = mass of the particle
We have chosen the expression such that x = 0, when t= 0.
 
    Velocity = dx/dt = Aω  Cos ωt

In one oscillation of period ωt = 2π, the velocity = 0 at  ωt = π/2 & 3π/2.
  Velocity is maximum at ωt = 0 ie., t =0, π, 2π ...

Displacement x = 0  when t = 0  and ωt = π, 2π..
Magnitude of x is maximum when  ωt = π/2  , 3π/2..

Total Mechanical energy in the system of the particle :
E = PE + KE = 0 + 1/2 m v²
   = 1/2 m A²ω² Cos² ωt     at x = t = 0.
   = 1/2 m A² ω² (1 + cos 2ωt) /2
   = 1/4  m A² ω² + 1/4 m A² ω²  cos 2ωt   --- (1)

   The total energy remains constant as there is no external force acting on the system (spring + mass or pendulum + string + gravity). The force acting on the system is conservative. So energy is conserved and transforms between PE and KE.

Average of a cosine wave over a period of ωt = 0 to 2π is zero as cosine wave is positive and negative equally during this period.

So Average energy = 1/4 m A² ω² = 1/4 m A² 4π² f²
                                 = π² m A² f²
where f = frequency of oscillation in Hz